Problem: Simplify; express your answer in exponential form. Assume $r\neq 0, z\neq 0$. $\dfrac{{(r^{-4})^{-2}}}{{(r^{4}z^{-3})^{-5}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${r^{-4}}$ to the exponent ${-2}$ . Now ${-4 \times -2 = 8}$ , so ${(r^{-4})^{-2} = r^{8}}$ In the denominator, we can use the distributive property of exponents. ${(r^{4}z^{-3})^{-5} = (r^{4})^{-5}(z^{-3})^{-5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r^{-4})^{-2}}}{{(r^{4}z^{-3})^{-5}}} = \dfrac{{r^{8}}}{{r^{-20}z^{15}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{8}}}{{r^{-20}z^{15}}} = \dfrac{{r^{8}}}{{r^{-20}}} \cdot \dfrac{{1}}{{z^{15}}} = r^{{8} - {(-20)}} \cdot z^{- {15}} = r^{28}z^{-15}$.